Covalent Bonding and Molecular Geometry Objective The objective of this exercise is to help in understanding the geometric relationships of atoms in simple molecules and the relationship of hybridization to the geometry present. Discussion In the last 30 years, data obtained from spectrometric measurements, Xray and electron diffraction studies, and other experiments have yielded precise information about bond distances, angles, and energies. In many cases, the data confirmed conclusions reached earlier. In other cases, valuable new insights were acquired.
Structure theory has advanced far beyond the simple electron dot representations and now rests securely on the foundations of quantum and wave mechanics. Although problems involving only simple molecules can now be solved with mathematical rigor, approximations such as the valence bond theory and the molecular orbital theory are very successful in giving results that agree with experimental measurements. This exercise will use valence bond theory or hybridization to look at the geometry formed from various hybridizations. You will use a framework model kit which gives the correct angles for the each of these hybridizations. The first bond formed between any two atoms is always a sigma (s) bond (one that is symmetric about the bond axis).
Additional bonds between the same two atoms will be pi (p) bonds (perpendicular to the bond axis).
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It is the sigmabonds and any lone pairs of electrons occupying the sigma hybrid orbitals that determine the geometry of a molecule. Pibonds are always perpendicular to the sigmabonds and follow the geometry formed by the sigma bonding. Procedure Check out a molecular model kit from the stockroom. Read the kit directions to see which framework center is used for each hybridization.
Tetrahedral (sp 3 hybridization) CH 4 Construct a model of meth an using a tetrahedral center (4 prongs) and four rods of the same color to show how the 4 H’s are attached. Geometry Lewis dot diagram # of s bonds on C Approximate H-C-H angel Max # atoms (incl. C) in one plane Is there a mirror plane (divides the molecule in equal halves) H 3 CCH 3 Construct a model of ethane using a tetrahedral center for each C and the same color rods for all 6 H’s with a CC bond present. Geometry Lewis dot diagram # of s bonds on each C Approximate H-C-H angle Approximate H-C-H angle The C-C bond is a single bond and has free rotation about it. Arrange the ethane molecule so that each CH bond on one C atom is exactly parallel to a CH bond on the second C atom. (This is the eclipsed position.
) View this arrangement by looking along the CC bond such that the atoms on the front C blank out those on the back C. From this view the model is represented as in I (a Newman projection. ) Now rotate the front carbon 60 degrees clockwise producing II. Which of the above arrangements allows more space per atom Which arrangement would you expect to be more stable CIH 2 CCH 2 Cl Make a model of 1, 2 dichloroethane by replacing an H atom with a C 1 atom on each C in your ethane model.
Use a different color rod to show the C 1 atoms. The eclipsed Newman projection is given in 111. Rotate the front C by 60 degrees and show the result on IV, rotate another 60 degrees and show the result on V, etc. , until you have completed a 360 degree rotation about the front C.
Remember that a C 1 atom is much larger than an H atom Which of the structures would you expect to be the most stable Explain why. Which of the structures would you expect to be the least stable Explain why. NH 3, Use a tetrahedral center for the N atom and three rods of the same color for the 3 H’s to make a model of ammonia. Geometry Lewis dot diagram # of s bonds on N # on non-bonding pairs on N Approximate H-N-H angle H 2 O Use a tetrahedral center for the O atom and two rods of the same color for the 2 H’s to make a model of water. Sketch the geometry Draw a dot diagram # of s bonds on O # on non-bonding pairs on O Approximate H-O-H angle Trigonal Planar (sp 2 hybridization) BF 3 Use a trigonal planar center (has 3 or 5 prongs depending on the model kit) and three rods of the same color for the three F’s to make a model of boron tri fluoride. (If you have a model kit without a 3 prong center, use the 5 prong center and put the three Fs on the 3 equivalent positions.
The Essay on Covalent Bond Atoms Groups Molecules
Bio-molecules are built by assembling atoms, by virtue of covalent bonds. The most collective elements that can be discovered in bio molecules are the CHOPS element, which stands for the basic abbreviation of carbon, hydrogen, nitrogen, phosphorus, and sulfur. Each one of these elements has a specific valence. A valence resolves the number from the covalent bonds it can be able to form. Presuming ...
) Geometry Lewis dot diagram # of s bonds on B # atoms in one plane Approximate F-B-F angle NO 31- The BF 3 model is the same model needed for the nitrate ion. Geometry Lewis dot diagram Total # of bonds on N# of s bonds on N# of p bonds on N If Px & Py are used for SP 2 hybridization what is the function of Pz Location of p bond compared to plane of the molecule Approximate O-N-O angle H 2 C = CH 2 Use sp 2 centers for both C atoms and four rods of the same color for the four H’s to make a model of ethene, which has a C = C bond. This is a planar molecule. If your model kit has a 5 prong center, align the 4 th and 5 th positions on the two C atoms parallel with each other. With a 3 prong center, put a rod through the hole in the center of each and align these rods parallel.
These represent the Pz orbital on each C atom which overlap and form the pibond between the carbons. In contrast to single bonds (sigma only) which have free rotation, there is no rotation around bonds that contain pibonds. You would have to break the pibond in order to rotate about such a bond and that requires too much energy for rotation to occur. (Rods through holes in center of 3 prong framework) (With the 3 prong centers, now that you realize there is no rotation about the C = C bond and why, remove the rods through the center holes to make it easier to see the planar nature of the molecule.
) Geometry Lewis dot diagram # of s bonds on C# of p bonds on C # atoms of ethene in one plane Approximate H-C-H angle Approximate H-C-C angle Consider the H numbers in the drawing above. Replace H 1 and H 3 with different color rods to represent two Cl atoms (cis-geometry. ) Make a second model of ethene and replace H 2 and H 3 with two Cl’s (trans-geometry. ) Are the cis-and trans-models the same If you replaced H 1 and H 4 with Cl’s, would it be the cis-, the trans- or another geometry If you replaced H 2 and H 4 what is the geometry If you replaced H 1 and H 2 what is the geometry Draw planar structural diagrams for each of the following molecules. Your drawing should show correct bond angles. H 2 C = CH 2 H 2 C = C HCl cis-HCIC = CClH trans-HCIC-CClH C 12 C = CH 2 Linear (sp hybridization) BeH 2.
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Use the center with 6 prongs for the Be atom and put two rods of the same color opposite each other then make a model of beryllium di hydride. Geometry Lewis dot diagram # of s bonds on Be# of non bonding pairs on Be # atoms in one plane Approximate H-Be-H angl Use two of the 6 prong centers for the carbon atoms and two same color rods for the H’s to make a model of ethyne. This is a linear molecule. Geometry Lewis dot diagram # of s bonds on C# of p bonds on C# atoms in one plane Approximate H-C-C angle If bond axis is x, what are along y and z Is there a free rotation along C-C bond Why Octahedral (sp 3 d 2 hybridization) SF 6 Use the 6 prong center for the S atom and rods of the same color for the 6 F’s to make a model of sulfur hexafluoride. Geometry Lewis dot diagram # of s bonds on S# of p bonds on S # of lone pairs on S Approximate F-S-F angle BrF 5 Remove one of the rods from the SF 6 model to make a model for bromine pentafluoride. Geometry Lewis dot diagram # of s bonds on Br# of p bonds on Br # of lone pairs on Br Approximate F-Br-F angle XeF 4.
Use a 6 prong center for the Xe and rods of the same color for the 4 F’s. Place all the F’s in the same plane around the Xe to make a model of xenon tetrafluoride. Geometry Lewis dot diagram # of s bonds on Xe# of p bonds on Xe # of lone pairs on Xe Approximate F-Xe-F angle Lone pair to lone pair angle Why are they opposite, not adjacent Trigonal Bi pyramid (sp 3 d hybridization) PF 5 Use the 5 prong center if your set has one and put five same color rods on for the F’s to make a model of phosphorous pentafluoride. If your set does not have a 5 prong center, use the 3 prong center and put a rod through the hole in the middle to make the 4 th and 5 th positions.
Geometry Lewis dot diagram # of s bonds on P# of p bonds on S # of lone pairs on P Approximate 1-P-F 3 angle Approximate F 3-P-F 5 angle SF 4 Remove the F 3 rod from the PF 5 model to make a model of sulfur tetrafluoride. Geometry Lewis dot diagram # of s bonds on S# of p bonds on S # of lone pairs on S Approximate F-S-F angle Why are two lone pairs in equatorial position Summary When we consider the models we have made, we see that they all obey the VESPER rules (see your textbook).
The Term Paper on What is the ‘covering law’ model of explanation?
Carl Hempel’s “covering law” model of explanation states essentially that an explanation for an event can be drawn from a set of general laws or, in the case of the social sciences, universal hypotheses. Hempel claims the study of history is not generally associated with the search for general laws governing historical events. However, history is a discipline within which the theory of “covering ...
These are summarized in the table below. Total # bonds plus lone pairs Hybridization on central atom Bond angle Our examples 2 sp 180 o BeH 2 HC = CH 3 sp 2 120 o BF 3, NO 31-, H 2 C = CH 2 4 sp 3 109. 5 o CH 4. NH 3, H 2 O H 3 CCH 3, H 2 CICCClH 2 5 sp 3 d 90, 120 o PF 5, SF 4 6 sp 3 d 2 900 SF 6, BrF 5, XeF 4.
